How to evaluate the given series ?

Question

2/2⁽²⁾+3/2⁽³⁾+4/2⁽⁴⁾ +.......

Answer 1

        $S  = \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + - - -$

$\frac{1}{2} \ast S =$           $\frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + - - -$

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Perform Substraction,

$\frac{1}{2} \ast S = \frac{2}{2^2} + \frac{1}{2^3}  + \frac{1}{2^4} + \frac{1}{2^5} + - - - - - $

$\frac{1}{2} \ast S = \frac{2}{2^2} + \frac{1/2^3}{1-1/2}$

$\frac{1}{2} \ast S = \frac{2}{2^2} + \frac{1/2^3}{1/2}$

$\frac{1}{2} \ast S = \frac{2}{2^2} + \frac{1}{2^2}$

$\frac{1}{2} \ast S = \frac{3}{2^2}$

$S = 2 \ast  \frac{3}{2^2} = \frac{3}{2}$

Note: for any such series having AP and GP in it this procedure will work.

Comments

Actually I was solving for this recurrence which has n at its end so then how to proceed ?

https://gateoverflow.in/34569/how-to-solve-below-recurrence-relation

Answer 2

Taking 1/2^2 common  we have  S = 1/2^2*(2+3/2+4/2^2.....) ,  Let us consider S1 = 2+3/2+4/2^2 .... , this is a AGP , so using AGP formula it is  S = 6